Take b = 2.

b^(n-1) mod n = 1.

67 is prime.

b^((n-1)/67)-1 mod n = 5486, which is a unit, inverse 7875.

3 is prime.

b^((n-1)/3)-1 mod n = 10181, which is a unit, inverse 3574.

(3 * 67) divides n-1.

(3 * 67)^2 > n.

n is prime by Pocklington's theorem.